Question

Find the sum

$(4 - \frac{1}{n} ) + (4 - \frac{2}{n} ) + (4 - \frac{3}{n} ) + ... \: upto \: \: n \: \: terms.\:$

Class 10

Arithmetic Progressions


⭕Best answer required ⭕

All the best

Answer 1

Answer

Given that,

$\implies ( 4 - \frac{1}{n} ) + ( 4 - \frac{2}{n} ) + ( 4 - \frac{3}{n} ) + .... \:\: n$

Taking Common terms outside we get,

$\implies ( 4 + 4 + 4 + .... \:\: n) - \frac{1}{n} ( 1 + 2 + 3 + ...  \:\: n ) \\\\\text{ Solving the first part we get,} \\\\\implies a = 4, \:\:\: d = 0, \:\:\: n = n \\\\\implies a_n = a + ( n - 1 )d \\\\\implies a_n = 4 + ( n - 1 ) 0 \implies 4 + 0 = 4\\\\S_n = \dfrac{n}{2} [ a + a_n ]\\\\\implies S_n = \dfrac{n}{2} [ 4 + 4 ] \implies \dfrac{8n}{2} \implies 4n$

Solving the next part we get,

$\text{ Let us solve only the inside part and then multiply them with } \:\: \dfrac{1}{n}\\\\ \implies a = 1, \:\:\: d = 1, \:\:\: n = n \\\\ \implies a_n = 1 + ( n - 1 ) 1 \\\\ \implies a_n = 1 + n - 1 \implies n\\\\ \implies S_n = \dfrac{n}{2} [ 1 + n ] \implies [ \dfrac{n^2 + n}{2} ] \\\\ \text{Substituting back in the equation we get,}\\\\ \implies 4n - \dfrac{1}{n} [ \dfrac{n^2+n}{2} ]\\\\\implies 4n - \dfrac{n^2 + n }{2n} \\\\ \implies 4n - \dfrac{n( n +1)}{n(2)} \\\\ \text{n gets cancelled and we get, } \\\\ \implies 4n - \dfrac{n+1}{2}$

$\text{ Taking LCM we get,} \\\\\implies \dfrac{ 8n - ( n + 1) }{2} \implies \dfrac{7n - 1}{2}$

Hence the sum of all terms is 7n - 1 / 2

Answer 2

$\underline{\underline{\mathfrak{\large{Solution : }}}}$




$\underline{\textsf{To Find : }}$




$\mathsf{ = \left(4 \: - \: \dfrac{1}{n} \right) \: + \left(4 \: - \: \dfrac{2}{n} \right) + ..... \: upto \: n \: terms \:}$




$\\ \mathsf{ = 4 \: - \: \dfrac{1}{n} \: + 4 \: - \: \dfrac{2}{n} + ..... \: upto \: n \: terms \:}$




$\\ \mathsf{ =(4 \: + \: 4 \: + .... \: upto \: n \: terms) \: - \: \dfrac{1}{n} \: - \: \dfrac{2}{n} ...... \: upto \: n \: terms }$




$\\ \mathsf{ = 4n \: - \: \left ( \dfrac{1}{n} \: + \: \dfrac{2}{n} \: + \: ....... \: upto \: n \: terms \right)}$




$\underline{\textsf{For Second Part : }}$




$\mathsf{\implies First \: term(a) \: = \: \dfrac{1}{n} } \\ \\ \\ \mathsf{\implies Common \: difference(a) \: = \: \dfrac{2}{n} \: - \: \dfrac{1}{n}} \\ \\ \mathsf{ \qquad \qquad \qquad \qquad \qquad \qquad \: = \: \dfrac{2 \: - \: 1}{n}} \\ \\ \mathsf{\qquad \qquad \qquad \qquad \qquad \qquad \: = \: \dfrac{1}{n} } \\ \\ \\ \mathsf{\implies No. \: of \: terms (N) \: = \: n }$




$\underline{\textsf{Using Formula : }}$




$\boxed{\mathsf{\implies S_{n} \: = \: \dfrac{n}{2}[2a \: + \: ( n \: - \: 1)d]}}$




$\\ \mathsf{\implies S_{n} \: = \: \dfrac{n}{2}\Bigg\{ 2 \: \times \: \dfrac{1}{n} \: + \: (n \: - \: 1)\dfrac{1}{n}\Bigg \}}$




$\\ \mathsf{\implies S_{n} \: = \: \dfrac{n}{2} \Bigg\{ \dfrac{2}{n} \: + \: \left( \cancel{n} \: \times \: \dfrac{1}{ \cancel{n}} \right) \: - \: \dfrac{1}{n}\Bigg\}}$




$\mathsf{\implies S_{n} \: = \: \dfrac{n}{2}\Bigg\{ \dfrac{2}{n} \: + \: 1 \: - \: \dfrac{1}{n} \Bigg\}}$




$\mathsf{\implies S_{n} \: = \: \dfrac{ \cancel{n}}{2}\Bigg\{ \dfrac{2 \: + \: n \: - \: 1}{ \cancel{n}} \Bigg\}}$




$\mathsf{\implies S_{n} \: = \: \dfrac{( n \: + \: 1)}{2}} \\ \\ \\ \textsf{Plug this value for Second Part }$

$\underline{\mathsf{Now,} }\\ \\ \\ \mathsf{= \: 4n \: - \: \Bigg( \dfrac{n\: + \: 1}{2} \Bigg)} \\ \\ \\ \mathsf{= \: \dfrac{ 8n \: - \: n\: - \: 1}{2}} \\ \\ \\ \mathsf{= \: \dfrac{7n \: - \: 1}{2}}$




$\mathsf{The \: required \: answer \: is \: \: \dfrac{7n \: - \: 1}{2}.}$